Lecture 6: Thermodynamic Potentials

3.020 Lecture 6 1 Thermodynamic potentials and equilibrium 1. For fixed U , the entropy S is maximized at equilibrium. 2. For fixed S and V , the internal energy U is minimized at equilibrium. 3. For fixed S and P , the enthalpy H = U + P V is minimized at equilibrium. 4. For fixed T and V , the Helmholtz free energy F = U −T S is minimized at equilibrium. 5. For fixed T and P , the Gibbs free energy G = U −T S+P V is minimized at equilibrium 2 Math note (aka fun with multi) State functions and independent variables • In 3.020, state functions are functions of most 3 independent variables. e.g. |{z} dU = T dS − P dV + µdN implies U = U (S, V, N ) We may never know this state function for most materials but it’s still a useful concept. • Thermodynamic potentials have “natural” independent variables S = S(U, V ) F = F (T, V ) U = U (S, V ) G = G(T, P ) H = H(S, P ) not writing “N” dependence to be compact 1 • Thermodynamic potentials are related by a change of variables via Legendre transforms U (S, V ) ⇐⇒ H(S, P ) | {z } exchanging V for P, H = U + PV beyond scope of 3.020 3 Coefficient relations and independent variables e.g. dU = T dS − P dV + µi dNi coefficients: T, P, µi independent variables: S, V, Ni ∂U ∂S | 4 ∂U ∂V = T, V,Ni = −P, S,Ni {z ∂U ∂Ni S,V,Nj 6=i coefficient relations = µi } General strategy for deriving thermodynamic relations Problem: Given dependent state variable Z and independent state variables (X, Y ), calculate ΔZ for a given process. e.g. Find ΔV for a process involving a pressure change at fixed entropy (i.e. a reversible adiabatic process) dV = |{z} X dP + |{z} Y dS find these coefficients 1. Write exact differential dV = ∂V ∂P dP + S 2 ∂V ∂S dS P Using X and Y as shorthand for unknown partial derivatives 2. Use DeHoff Table 4.5 to express dP, dS in terms of dP, dT dP = |{z} dT + |{z} dP 0 trivial 1 CP dT − V αdP T CP dV = XdP + Y ( dT − V αdP ) T dS = 3. Collect terms dV = Y CP dT + (X − Y V α)dP T 4. Compare to known coefficients dV = V αdT − V βdP ‘from Table 4.5’ and equate ∂V ∂T ∂V ∂T =Y P CP =Vα T = X − Y V α = −V β P 5. Solve for unknowns X, Y Y = =⇒ V αT , CP X= V 2 α2 T −Vβ CP V 2 α2 T V αT dV = ( − V β)dP + dS CP CP | {z } integrate this to find change ΔV for pressure change at fixed entropy 3 for ideal gas, simplifies to X=− leading to V CP − R P CP Pf inal Vinitial γ =( ) , Vf inal Pinitial γ= CP CV in PSet 2, claculate dG = |{z} dP + |{z} dS 5 Equilibrium at fixed T and P G = U + P V − T S is minimized dG = dU + P dV + V dP − T dS − SdT ← chain rule = T dS − P dV + µi dNi + P dV + V dP − T dS − SdT = −SdT + V dP + µi dNi ← combined statement • independent variables: T, P, Ni • coefficients: −S, V, µi • by inspection: ∂G ∂T P,Ni ∂G V = ∂P T,Ni ∂G µi = ∂Ni T,P,Nj6=i −S = • at fixed T and P , dT = 0 and dP = 0, and equilibrium condition reduces to dG = −S |{z} dT +V |{z} dP +µi dNi = 0 0 0 because G is minimized at equilibrium 4 • for multicomponent, heterogeneous systems, this becomes dG = µki dNik superscript k = phase label subscript i = component label • evaluating equilibrium requires knowing how components can exchange between phases, and the effect on the µki s • About independent (“natural”) variables – If a variable is regulated, a.k.a held constant, then it should be independent – Identifying independent & dependent variables from problem statements is KEY to success at thermodynamics e.g. “... at fixed pressure...” −→ choose P as one independent variable e.g. “...adiabatic process...” −→ choose S as one independent variable e.g. “...with temperature held constant...” −→ choose T as one independent variable etc. 5